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September 15th, 2016, 06:24 PM  #1 
Newbie Joined: Sep 2016 From: CO Posts: 6 Thanks: 0  Solve this recurrence relation with repeated subsitution
$\displaystyle \sqrt(n)T(\sqrt(n)+n$ I've tried what I think is right, but I came up with: $\displaystyle (n^{1/2}*n^{1/2^{2}}*...*n^{1/2^{k}})T(n^{1/2^{k}})+(n^{1/2^{k1}}+...+n^{1/2}+n)$ And I have no idea how to find the pattern of that so I'm hoping I'm wrong. 
September 15th, 2016, 06:49 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
What recurrence relation?! You've given an expression in $n$. Why did you post this in Linear Algebra?

September 15th, 2016, 07:27 PM  #3 
Newbie Joined: Sep 2016 From: CO Posts: 6 Thanks: 0 
This is for my Analysis of Algorithms class. I'm not sure where else to put it so I was hoping Linear was related.

September 15th, 2016, 07:28 PM  #4 
Newbie Joined: Sep 2016 From: CO Posts: 6 Thanks: 0 
As for the format, does this change anything? $\displaystyle T(n) = \sqrt(n)T\sqrt(n)+n$ for $\displaystyle n\leq 2$ 
September 15th, 2016, 07:38 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
Um...shouldn't the $n$ have a subscript?

September 15th, 2016, 07:58 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
The system caught your link. I've approved the post. Just to be sure we're on the same page, do you intend $$T(n)=\sqrt nT(\sqrt n)+n$$ ? 
September 15th, 2016, 08:04 PM  #8 
Newbie Joined: Sep 2016 From: CO Posts: 6 Thanks: 0 
Yes, sorry I'm not good at Latex yet.

September 16th, 2016, 03:03 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
From WA: $$T(n)=\frac{n}{e^2}+\frac{n\log(\log(n))}{\log(2) }$$ I've verified it with a CAS and I don't know how to compute it. It seems like a somewhat (at least) difficult problem. 
September 16th, 2016, 03:35 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
The OP's relation is invalid for n=1. It also fails to define T for most values of n as far as I can see.


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recurrence, relation, repeated, solve, subsitution 
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